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NT
0
HV
2
KN
23 tháng 7 2019
\(C=10^2+8^2+6^2...+2^2-\left(9^2+7^2+5^2+...+1^2\right)\)
\(\Rightarrow C=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(\Rightarrow C=\left(10-9\right)\left(10+9\right)+\left(8-7\right)\left(8+7\right)+...+\left(2-1\right)\left(2+1\right)\)
\(\Rightarrow C=19+15+...+3\)
Vậy C = {(19 + 3)[(19-3):4+1]} :2 = 60
23 tháng 7 2019
\(C=10^2-9^2+8^2-...-3^2+2^2-1^2\)
\(=\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=10+9+8+...+2+1\)
\(=\frac{\left(1+10\right)10}{2}=55\)
Vậy C=55
MN
2
26 tháng 9 2018
= \(10^2+8^2+6^2+4^2+2^2-9^2-7^2-5^2-3^2-1\)-1
=\(55\)
30 tháng 10 2018
\(\left(10^2+8^2+6^2+4^2+2^2\right)-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(=10^2+8^2+6^2+4^2+2^2-9^2-7^2-5^2-3^2-1^2\)
\(=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(10-9\right)\left(10+9\right)+\left(8-7\right)\left(8+7\right)+...+\left(2-1\right)\left(1+2\right)\)
\(=10+9+8+7+...+2+1\)
\(=\frac{\left(1+10\right)\cdot10}{2}\)
\(=55\)
KT
1
10 tháng 10 2017
\(\left(10^{12}+5^{11}.2^9-5^{13}.2^8\right):4.5^5.10^6\)
\(=\left(5^{12}.2^{12}+5^{11}.2^9-5^{13}.2^8\right):2^2.5^5.2^6.5^6\)
\(=\left[5^{11}.2^8\left(5.2^4+2-5^2\right)\right]:2^8.5^{11}\)
\(=\frac{5^{11}.2^8\left(5.16+2-25\right)}{5^{11}.2^8}\)
\(=80+2-25\)
\(=57\)
NT
4
CL
22 tháng 6 2015
A = 232 + ( 223 + 223 - 224 ) + ( 218 - 217 - 217 ) + ( 29 + 29 - 210 )
A = 232 + 1 (các biểu thức trong dấu ngoặc đều bằng 0).
HN
1 tháng 1 2018
a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)
\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)
c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)
\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)
\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{5}{2}\)
e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)
\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)
\(=\dfrac{-3\left(x-13\right)}{2x^3}\)
g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).
19 tháng 11 2017
Ta có: P = 102 - 92 + 82 - 72 + ...+ 22 - 1
= (102 - 92) + (82 - 72) + ... + (22 - 1)
= (10 + 9)(10 - 9) + (8 + 7)(8 - 7) + ... + (2 - 1)(2 + 1)
= 19 + 15 + 11 + 7 + 3
= \(\dfrac{\left(19+3\right).5}{2}=55\)
20 tháng 11 2017
\(P=10^2-9^2+8^2-7^2+...+2^2-1\\=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\\ =\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2+1\right)\left(2-1\right)\\ =19+15+11+7+1\\ =53\)